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Small cases lie — page 8

Patterns that hold for every example anyone would check by hand, and then stop. The cases within reach are not a sample of the cases.
Malfatti's three circles against the three largest, one at a time. A triangle with angles 40, 70, 70 degrees: Malfatti's circles cover 69.1% of it, the greedy choice of incircle then largest remaining circles covers 73.7%. Geometry

Three circles that touch and are not the largest

In 1803 Gian Francesco Malfatti asked how to cut three round columns from a triangular block of marble with as little waste as possible, and answered with three circles each touching the other two and two sides. The circles exist in every triangle and are found by solving three equations. They are never the answer to his question: the plain greedy rule — the inscribed circle first, then the largest circle that still fits, twice — always does better, by 1.4 per cent on the equilateral triangle and by almost double on a thin one.

The tree writes 4/17 as four unit fractions, and a search finds three. Left-neighbour chain 4/17, 3/13, 2/9, 1/5, 0/1, gaps 1/221, 1/117, 1/45, 1/5; three-term decomposition 1/5 + 1/30 + 1/510. Number

Three unit fractions for every four over n

Two neighbouring fractions in the tree always differ by a unit fraction, so stepping down the tree writes any fraction as a sum of them — four over n in at most four steps. Erdős and Straus asked in 1948 whether three always suffice. Three identities settle every n except those leaving remainder 1 on division by 24, a search settles every one anyone has tried, and nobody has a proof.

Luxembourg's votes and Luxembourg's power, 1958 to 1995. Paired bars for Luxembourg at five enlargements of the Council: vote share falling and power share rising from nothing in 1958 to 0.95% in 1973 and 3.02% in 1981. Applied

A vote worth nothing until it was outnumbered

From 1958 to 1973 Luxembourg held one vote of seventeen in the Council of the European Communities and could never once change an outcome. When the Council grew, Luxembourg's share of the votes fell and its share of the power rose from nothing. Power is not a quantity a member holds; it is a property of the whole assembly, and changing the assembly moves it in directions nobody would guess.

Three kinds of member, and a triple that breaks a grouping. Two copies of one market with three members of each of three kinds and cyclic preferences: on the left a grouping blocked by the triple B, 1, p; on the right a stable grouping. Applied

A third kind of member

Two sides always have a stable matching, and the proof is a procedure. Add a third kind of member — capitals who rank numbers, numbers who rank letters, letters who rank capitals — and every market small enough to check still has a stable grouping. Ten million markets of three have at least ten each. Nobody can prove it continues, because the structure the two-sided proof stands on is gone.

The 11 trees on 7 points, and how many labellings each has. All trees on 7 points with their numbers of symmetries and of distinct labellings; the labellings add to 16807. Discrete

Almost every tree can be turned over

Cayley's n^(n−2) counts trees with labels on their points. Take the labels off and the count has no formula, because a symmetric shape absorbs labellings: the star on seven points can be labelled only seven ways, the one asymmetric shape 5,040. The bookkeeping that reconciles the two counts says something unexpected — almost every tree, labelled or not, can be turned over onto itself, where almost every graph cannot.

Seventy, and the totals its divisors can make. A strip of the totals 0 to 74 marking those reachable as sums of distinct proper divisors of 70; 70 itself and the excess 4 are not. Number

Abundant, and still not a sum of its parts

Seventy's proper divisors add to seventy-four, more than seventy itself — and yet no selection of them adds to exactly seventy. The reason is that the excess, four, cannot be made from the small divisors one, two and five. Numbers like seventy are called weird; there are thirty-six below twenty thousand, every one of them even, and nobody knows whether an odd one exists.

How often each procedure finds the right verdict, with 3 judges of competence 0.7. both true: liable: premises 61.5%, conclusion 48.5%; first only: not liable: premises 83.1%, conclusion 88.6%; second only: not liable: premises 83.1%, conclusion 88.6%; neither: not liable: premises 95.3%, conclusion 97.7%; average 80.7% against 80.9%. Applied

Which vote finds the truth more often

A court that must decide two premises and their conjunction can vote on the premises or on the conclusion, and the two votes sometimes disagree. Suppose there is a right answer and each judge is more often right than wrong. Then the question has an exact answer: for three judges the conclusion vote is more often right when competence is below √½ and the premise vote above it, and in a large court the premise vote wins at any competence — while the conclusion vote turns out to be a built-in standard of proof.

Records in a thousand draws, and the lengthening waits between them. A sequence of 1000 uniform draws on a logarithmic position axis, with records at positions 1, 6, 11, 27, 36, 145, 539, 927. Probability

A wait that ends and has no average

Draw a number, then keep drawing until one beats it. Half the time the very next draw does. The wait is certain to end, it is usually short, and its average is infinite — and the chance of still waiting after n draws is exactly 1/(n + 1), whatever the numbers are drawn from, because symmetry is the only thing the calculation uses.

Running totals of random steps, stopped as they pass 4. Four staircases of uniform steps stopped on passing 4, taking 11, 8, 8, 10 steps; the exact mean is 8.666604. Probability

Two-thirds of a step past the line

Add random numbers between nought and one until the total passes one, and it takes e of them on average. Ask the same about passing any t and the answer is an exact finite sum that settles, within a few units, onto the line 2t + 2/3. The slope is one over the average step. The two-thirds is the interesting part: it is what the one step that crosses the line charges for being chosen by its length, and it is set entirely by how much the steps vary.

The 3,600 pairs of icosahedral rotations, and what each pair generates. A 60 by 60 grid of ordered pairs of elements of A5 shaded by the order of the subgroup they generate; 2280 generate A5. Algebra

Nineteen copies and no more

Two of the icosahedron's rotations, chosen at random, generate all sixty of them nineteen times in thirty. Philip Hall found the number in 1936 by inclusion and exclusion over every subgroup at once, weighted by the Möbius function of the subgroups' order — and the same count says that nineteen copies of the group, side by side, can still be generated by two elements, while twenty cannot.

Products of widening sinc waves, all with area π/2 but the last. Graphs of the products of sinc(x/(2k+1)) for k = 0 to n, n = 0, 1, 3, 7, on 0 ≤ x ≤ 24. Analysis

Seven times half of π, and then not

The area under sin x / x, out to infinity, is π/2. Multiply in sin(x/3)/(x/3) and the area is still π/2; multiply in the waves for 5, 7, 9, 11 and 13 and it is still π/2, exactly. Add the wave for 15 and the area falls short — by about 2.3 × 10⁻¹¹. Nothing about the curves changes at the eighth factor. What changes is that 1/3 + 1/5 + … + 1/15 has just passed 1.

How good each judge must be, as the premises multiply. k = 1: 0.5000; k = 2: 0.7071; k = 3: 0.7937; k = 4: 0.8409; k = 5: 0.8706; k = 6: 0.8909; k = 7: 0.9057; k = 8: 0.9170; k = 9: 0.9259; k = 10: 0.9330. Applied

Every premise makes the verdict vote stricter

A court that must find several things true before it finds liability can vote on each of them or on the verdict. With two elements, judges right seven times in ten are enough for either. With five, the verdict vote needs judges right eighty-seven times in a hundred, or a larger court makes it worse — while the premise vote still needs only better than even. And when any one ground is enough instead of all, the two procedures swap their failures.

Seven towns joined with junctions and without. Seven random points: spanning tree 1.65407, Steiner tree 1.54997 with 2 junctions, ratio 0.93707. Geometry

What a junction can save

Join a set of towns by roads that may meet only at towns, and the shortest such network is easy to find. Allow junctions anywhere and the network gets shorter — but never, it is conjectured, by more than 13.4%, the saving the Fermat point makes on an equilateral triangle. Computed exactly on 1,304 random sets of up to seven towns, the average saving is about 3%. An adversarial search drives it down to the bound and no further, and a hexagon with its centre attains it exactly. The proof of the bound that was accepted for twenty years has a gap.

Primes in five thousand thin sectors. 581517 primes in (2e7, 4e7] in 4967 sectors; mean 117.08, variance ratio 0.874, 2 empty. Number

Sectors that shrink with the primes

Hecke proved that the angles of the primes a² + b² are evenly spread, so every fixed sector eventually gets its share. Let the sector shrink as the primes grow, to width X^(−α), and the question is open beyond small α. Measured on the 581,517 such primes between twenty and forty million, wide shrinking sectors fluctuate far less than chance and thin ones as much as chance — and the thin sectors that go empty are not scattered: they gather beside the directions of small rational slope, most of all slopes like 1/3 and 1/1 whose two terms are both odd.

The longest run of heads at three sizes, never the same shape. n=65536: 12:0.018 13:0.117 14:0.233 15:0.239 16:0.172 17:0.104 18:0.057 19:0.030 20:0.015; n=82273: 13:0.075 14:0.204 15:0.249 16:0.197 17:0.124 18:0.070 19:0.037 20:0.019; n=104408: 13:0.040 14:0.162 15:0.248 16:0.221 17:0.148 18:0.086 19:0.046 20:0.024 21:0.012. Probability

The longest run has no limit

Toss a fair coin n times and the longest run of heads is about log₂ n. The average settles: it is log₂ n − 0.667 to within a few millionths by a million tosses. The distribution does not. Because the run is a whole number and log₂ n is not, the chance of each value swings round the same cycle once every doubling of n, for ever, and computing it exactly to a million tosses shows the swing as large at the end as at the start.

The share of numbers beginning with 1, never settling. 10^1.0: 0.2000, 10^1.8: 0.1746, 10^2.3: 0.5487, 10^2.8: 0.1841, 10^3.3: 0.5231, 10^3.8: 0.1931, 10^4.3: 0.5002, 10^4.7: 0.2022, 10^5.2: 0.4766, 10^5.7: 0.2117, 10^6.2: 0.4519, 10^6.7: 0.2217, 10^7.2: 0.4261, 10^7.7: 0.2321, 10^8.2: 0.3990, 10^8.7: 0.2431, 10^9.2: 0.3707, 10^9.6: 0.2545, 10^10.1: 0.3411, 10^10.6: 0.2665, 10^11.1: 0.3100, 10^11.6: 0.2791. Probability

A share that depends on the average

What share of the whole numbers begin with the digit 1? Counted up to N, the answer swings between one ninth and five ninths for ever, and averaging the swing over N, even three times over, narrows it without removing it. Weight each number n by 1/n instead and the share settles at log₁₀ 2 — Benford's value — and so does every other sensible weighting that counts each decade alike. The primes behave the same way. Benford's law for the counting numbers is not a fact about them; it is a fact about how they are averaged.

Every cubic x³ + bx + c up to 30, sorted by its Galois group. A 61 by 61 grid of cubics x³ + bx + c: 309 reducible, lying on lines c = −r³ − rb; 20 cyclic with square discriminant; 3392 with group S3. Algebra

Twenty-seven million cubics, sorted by their symmetry

Write down every cubic x³ + ax² + bx + c with coefficients up to 150 and sort each by its Galois group. 98.7% have all six symmetries of three roots. The rest are reducible at a rate of fifteen in every H², or cyclic at five in every H to the three-halves — and the cyclic ones turn up four times as often as chance would allow, because discriminants are not random numbers.

Even quartics never have the full group, and one step away almost all do. Even quartics x^4 + bx^2 + d to 20: reducible 199, S4 0, A4 0, D4 1338, C4 20, V4 124. With +x: reducible 157, S4 1524, A4 0, D4 0, C4 0, V4 0. Algebra

The quartics whose roots stay paired

A quartic can have five different Galois groups, and in a box of nearly fourteen million quartics they do not come in order of size. The eight-element group D₄ outnumbers the twelve-element A₄ thirty-five to one. Every even quartic has a smaller group than S₄, and moving one step off that plane restores the full group almost everywhere.

The sum of all factorials in base 5: digits that stop changing. Partial sums of n! for N = 0 to 21, last 14 base-5 digits; the 5-adic limit ends …13010230042224. Analysis

A sum of factorials that converges at every prime

1 + 1 + 2 + 6 + 24 + 120 + … grows faster than any geometric series and has no sum in the real numbers. Measure size by divisibility instead and the same series converges at every prime at once, to a different number each time. Whether any of those numbers ends in zero is a question Đuro Kurepa asked in 1971, and a search through every prime below 100,000 says no while a coin-toss model says it should have said yes about twice.

A chromatic polynomial with complex roots and log-concave coefficients. Wheel W5: P(x) = x⁶ − 10x⁵ + 40x⁴ − 80x³ + 79x² − 30x; |coefficients| 30, 79, 80, 40, 10, 1; min ratio 2.0000; roots 3.000, 0.000, 2.000−1.000i, 2.000, 1.000, 2.000+1.000i. Discrete

Log-concave with nothing to make it so

The chromatic polynomial of a graph counts its colourings, and its coefficients rise to one peak and fall, each squared at least the product of its neighbours. For a real-rooted polynomial that would be automatic. But by nine vertices fewer than one graph in twenty has only real chromatic roots, and the coefficients stay log-concave anyway — on every graph tested, and, by June Huh's theorem of 2012, on every graph there is.

Two random strings of bits and the longest thread between them. Two random binary strings of length 36; longest common subsequence 27. Probability

The longest thread between two random strings

Write down two random strings of a thousand bits each and find the longest sequence that can be read in both, skipping freely. It is about 81% of each, and the exact share — the Chvátal–Sankoff constant — has resisted computation for fifty years. Its fluctuations are a smaller mystery with a sharper edge: the variance is known to be at most n/2, and over every length that can be computed it grows like n to the power 0.7, with nobody able to say what it does after that.

The primes p for which Euclid's number p# ± 1 is itself prime. p up to 1100: p# + 1 prime for 2, 3, 5, 7, 11, 31, 379, 1019, 1021; p# − 1 prime for 3, 5, 11, 13, 41, 89, 317, 337, 991. Number

When Euclid's number is itself prime

Multiply the primes up to p and add one. Euclid needed only that the result has a prime factor beyond p; it is itself prime for p = 2, 3, 5, 7 and 11, and then not again until 31, and then not until 379. Up to 1,100 there are nine such primes of each sign, against thirteen that a careful estimate expects. The estimate predicts infinitely many and grows by about 1.8 every time x is multiplied by e; nobody can prove there is even one more.

The first ten Carmichael numbers and the divisions that make them. 561 = 3·11·17; 1105 = 5·13·17; 1729 = 7·13·19; 2465 = 5·17·29; 2821 = 7·13·31; 6601 = 7·23·41; 8911 = 7·19·67; 10585 = 5·29·73; 15841 = 7·31·73; 29341 = 13·37·61. Number

The numbers that fool every base

561 = 3 · 11 · 17 passes Fermat's test for every base it shares no factor with, and so do 1,105, 1,729 and 102 more numbers below ten million. They were proved infinite only in 1994. Counted to 10²¹ there are about twenty million, growing like x to a power that has crept from 0.21 to 0.35 — just past the bound the proofs guarantee and nowhere near the power of one that Erdős's argument predicts.

Two stable cycles a hair from the classical Hénon attractor. The Hénon attractor at a = 1.4 with the certified stable period-13 orbit at a = 1.39945 and period-19 orbit at a = 1.4001616 drawn over it. Dynamics

Stable cycles a hair from Hénon's chaos

Nobody knows whether the Hénon map at a = 1.4 is chaotic, because a stable cycle in a window too thin to see would look exactly the same. So hunt the windows: a stable orbit of period 19 at a = 1.4001616, proved to exist and to attract by interval arithmetic, in a window 15 hundred-millionths wide and 0.00016 from the classical value — and nothing nearer, down to a spacing of a ten-billionth.

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